Suppose \(X\)  is not strictly convex and let \({\bf L}:=\lbrace \lambda x+(1-\lambda )y;\;\;0\le \lambda \le 1\rbrace \)  be a nondegenerate line on the norm's unit
sphere \(S\) . Consider distinct unit vectors \(z_1,z_2\)  from the relative interior of \({\bf L}\subseteq S\) .
Then for some \(\varepsilon >0\)  we have
\(\Vert (1-\lambda ) z_1+\lambda z_2\Vert  =1;\qquad \lambda \in (-\varepsilon ,1+\varepsilon ).\) 
Let now \(f\)  be an arbitrary supporting functional at \(z_1\) , that is, \({f(z_1) = \Vert  f \Vert  = 1}\) .
Then
\(|1 + \lambda (f(z_2)-1)|=|f((1-\lambda ) z_1+\lambda z_2)| \le \Vert f \Vert  \cdot \Vert  z_1 + \lambda (z_2 - z_1) \Vert  = 1\) 
for any \(\lambda \in (-\varepsilon , \varepsilon )\) , so \(f(z_2)=1\) .
Hence \(f\)  is a supporting functional for both \(z_1\)  and \(z_2\) , and thus the sets of supporting functionals at \(z_1\) 
and at \(z_2\)  coincide. This implies that the set \(\lbrace x\in X;\;\;z_1\perp x\rbrace \)  coincides with the set
\(\lbrace w\in X;\;\;z_2\perp w\rbrace \)  because, by [1]}, \(z \perp x\)  if and only if \(x\)  is annihilated by some
supporting functional at \(z\) . As such, also \(N_{z_1}=N_{z_2}\) .
To prove the converse we assume \(X\)  is strictly convex. We need to show that for any two distinct vertices
\(z_1, z_2 \in \hat{\Gamma }(X)\)  we have \(N_{z_1} \ne N_{z_2}\) .
Note first that \(x \perp y\)  in \(X\)  if and only if \(x \perp y\)  in \(\mathop {\mathrm {span}}\lbrace  x, y \rbrace  \subseteq X\) , since,
by the definition, BJ orthogonality of two vectors depends only on the norms of their linear combinations. We
now pass to the subspace \(Y= \mathop {\mathrm {span}}\lbrace z_1, z_2\rbrace \) . Clearly, the norm on \(Y\)  is also strictly convex. It is
sufficient to show that \(N_{z_1} \ne N_{z_2}\)  in \(\hat{\Gamma }(Y)\) .
Consider some supporting functional \(f\)  at \(z_1\)  in \(Y\) , and let \(x \in \mathop {\mathrm {Ker}}f\) . Hence \(z_1 \perp x\) , so \(x\in N_{z_1}\) . Assume, to reach a contradiction, that \(x \in N_{z_2}\) . Then there exists a supporting functional
\(g\)  at \(z_2\)  in \(Y\)  such that \(x \in \mathop {\mathrm {Ker}}g\) .
Since \(\dim Y= 2\)  and \(\mathop {\mathrm {Ker}}f \cap \mathop {\mathrm {Ker}}g\)  is nontrivial, it follows that \(\mathop {\mathrm {Ker}}f = \mathop {\mathrm {Ker}}g\)  which implies that \(g\)  is a unimodular scalar multiple of \(f\) .
By transferring this scalar multiple to \(z_2\)  we may assume that \(f(z_1) =f(z_2) = \Vert f\Vert  = \Vert z_1\Vert  = \Vert z_2\Vert  = 1\) , so it follows that
\(1 = f(\lambda z_1 + (1 - \lambda ) z_2) \le \Vert f\Vert  \cdot \Vert \lambda z_1 + (1 - \lambda ) z_2\Vert  \le 1\) 
for all \(\lambda \in [0,1]\) . Thus the line segment \([z_1,z_2]\)  consists of unit vectors, a contradiction to
the strict convexity. Hence \(x \notin N_{z_2}\)  and \(N_{z_1} \ne N_{z_2}\) .
